UPSC NDA I 2018 Mathematics Solved Paper

Concept:For a quadratic with real coefficients, non-real complex roots occur in conjugate pairs. The sum of the roots is real, and the discriminant must be negative. Given that the real part of each root is 1, we can deduce the sum and then the condition on $\beta$.Explanation:Let the two distinct non-real roots be $1 + iy$ and $1 - iy$, where $y \neq 0$ and $y$ is real.The sum of the roots is $(1+iy) + (1-iy) = 2$.For the quadratic $z^2 + \alpha z + \beta = 0$, the sum of roots is $-\alpha$. Thus $-\alpha = 2$, so $\alpha = -2$.The product of the roots is $(1+iy)(1-iy) = 1 + y^2 = \beta$.Since $y^2 > 0$ (roots are distinct and non-real), we have $\beta > 1$.Alternatively, using discriminant: for non-real roots, discriminant $D = \alpha^2 - 4\beta < 0$. Substituting $\alpha = -2$ gives $4 - 4\beta < 0$, so $\beta > 1$.Therefore $\beta$ must lie in the interval $(1, \infty)$.Answer:$\beta \in (1, \infty)$